Title Crackiitjee.in.Phy.ch18 Physics Electricity Electrical Resistance And Conductance Electric Current Series And Parallel Circuits 1.6 MB 19
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Resistance

Resistance in Series:

“The Connected resistances are said to be in series when a potential

difference that is applied across the combination is the sum of the resulting

potential differences across the individual resistances".

The current through each resistor in series is necessarily the same but this is

not the sufficient condition for resistors in series. The condition becomes

sufficient when the current through each resistor is the same as the current

through the terminals of the network.

A combination of three resistances R1, R2 and R3 in series. The current through

each resistor is the same as the current i through the terminals A or B of the

networks. Suppose that V1, V2 and V3 are the potential differences across R1' R2

and R3 respectively. If V = VA - VB be the potential difference across the network,

then according to the definition of series combination.

V = V1 + V2 + V3

== iR1 + iR2 + iR3

== i (R1 + R2 + R3)

Important Points:

Total potential (V) is divided in such a way that :

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V1 : V2 : V3 == R1 : R2 : R3

Or,

If R1= R2 =R3 then V1= V2 = V3= V/3 and Req =3R

Similarly, if there are n resistors in series,

Req=R1 +R2+ .. + Rn

If R1= R2 =R3' then V1= V2 = V3= V/3 and Req =3R

, Similarly, if there are n resistors in series,

Req=R1 +R2+ .. + Rn

RESISTANCES IN PARALLEL

The connected resistances are said to be in parallel when a potential

difference that is applied across the combination is the same as' the

resulting potential difference across the individual resistances. It shows

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2. If nr » R, then, I =
R

nE
=

r
E (approx), that is, if internal resistance

of the connected cells is much greater than the external resistance,

then nearly the same current is obtained by single cell. Hence, there

is no advantage of connecting cells in series.

3. If in series grouping of n cells, n1 cells are reversed then

Eeq = (n- n1)E - n1E = (n - 2n1)E and req = nr

So I = (n - 2n1) E

R + nr

2. In Parallel: In this combination, the positive poles of all the cells are

connected to one point, and the negative poles to another point . Suppose

m cells, each of e.m.f. E and internal resistance r, are connected in parallel

and this battery of m cells is connected to an external resistance R. Since

the

cell are connected in parallel, the e.m.f. of the battery will also be E. If the

equivalent internal resistance of the cell be R1, then

R1
1

=
r
1

+
r
1

+ ..... upto m terms=
r
m

or R1 =
m
r

Therefore, Total resistance of the circuit = (r/m + R). If the current in the

external circuit be l,

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Then, l =
(r/m)+ R

E
=

r + mR
mE

Important Points:

i. If r<<R, that is, if the internal resistance of the cells is much

smaller than the external resistance, then r can be

neglected in comparison to mR. Then, from eq. (i), l = E/R

(approx.), i.e., total current will be equal to the current

given by a single cell. Hence, there is no advantage of

connecting the cells of small internal resistance in

parallel.

ii. If (r/m)>>R, that is, if the internal resistance of the cells in

larger than the external resistance, then the current will

be I =
R

nE (approx.). This current is nearly m times the

current given by r

a single cell. Hence, when the internal resistance of the

cell is much larger than the external resistance, then the

cells should be connected in parallel.

Mixed Grouping: - In this combination, a certain number of cells are

connected in series, and all such series. Combinations are then connected

mutually in parallel.

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Calculate the current in terms of Jo and the conductor's cross sectional area is

A = nR
2

(b) Suppose that instead the current density is a maximum J 0 at the surface and

decreases linearly to zero at the axis so that J = J = J0
R
r . Calculate the current.

Solution. (a) We consider a hollow cylinder of radius r and thickness dr.

The cross-sectional area of considered element is dA = 2πrdr

The current in considered element is

dI = JdA = J0 (1 -
R
r

)2rrdr

or dI = 2rJ0(1-
R
r

)rdr

Therefore, I = 2rJ0 (
0

R

# 1-
R
r

)rdr

I = I = J0
3
rR2

= J0
3
A

(b) I =
R

2rJ0

0

R

#
R
r2

dr

I =
R

2rJ0
R
r3; E

0

R

I =
R

2rJ0
R
r3

=
3

2A
J0

Example. A network of resistance is constructed with Rl & R2 as shown in the . The

potential at the points 1, 2,3,……….., N are V1, V2, V3,…….., VN respectively each

having a potential k time smaller than previous one. Find:

1)
R2
R1

and
R3
R2

in terms of K

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2) Current that passes through the resistance R2 nearest to the V0 in terms V0,

k and R3.

Solution.

1) According to kcL,

I = I1 + I2

Or
R1

V0 -
k
V0

=
R2

k
V0 -0

+
R1

k
V0 -

k2
V0

Or
kR1

(k - 1)V0
=

kR2
V0 +

k2R1

(k - 1)V0

Therefore,
R2
R1 =

k
(k -1) 2

Also, I

= I

1 + I

2

Or
R1

kN
-2

V0 -
kN

-1

V0

=
R2

kN
-1

V0 -0
+

R1 + R3
kN

-1

V0 -0

After Solving,

R3
R2 =

k - 1
k

2) Here, I1 =
R2

V1 -0 =
R2

k
V0 -0

I1 =
kR2
V0 =

k(
k - 1

k
)R3

V0 =
k2R3

(k - 1)V0

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